The electric potential V is given as a function of distance x (metre) by V=(5 x2+10 x-9) volt. Value of electric field at x=1 is

Question

The electric potential V is given as a function of distance x (metre) by V=(5 x2+10 x-9) volt. Value of electric field at x=1 is (A) 20 V/m (B) 6 V/m (C) 11 V/m (D) -23 V/m

Tardigrade Admin · Accepted Answer

E=-(d V/d x) =-(d/d x)(5 x2+10 x-9) =-10 x-10 ∴ (E)x=1=-10 × 1-10 =-20 V / m

Q. The electric potential $V$ is given as a function of distance $x$ (metre) by $V = (5 x^{2} + 10 x - 9)$ volt. Value of electric field at $x = 1$ is

20 V/m

6 V/m

11 V/m

-23 V/m

$E = - \frac{d V}{d x}$
$= - \frac{d}{d x} (5 x^{2} + 10 x - 9)$
$= - 10 x - 10$
$∴ (E)_{x = 1} = - 10 \times 1 - 10$
$= - 20 V / m$

Q. The electric potential V is given as a function of distance x (metre) by V=(5x2+10x−9) volt. Value of electric field at x=1 is